int
?GetMaxCountofContinuousLetter(
char
?
*
?p,
char
?searchChar)
{
????
int
?maxCount?
=
0
;
????
int
?currentCount?
=
?
0
;
????
while
(
1
)
????
{
????????
if
(
*
p?
==
?searchChar)
????????
{
????????????currentCount?
=
?
0
;
????????????
while
(
*
p?
&&
?
*
p?
==
?searchChar)
????????????
{
????????????????currentCount?
++
;
????????????????p
++
;
????????????}
????????????
if
(maxCount?
<
?currentCount)
????????????????maxCount?
=
?currentCount;
????????}
????????
if
(
*
p?
==
?
'
\0
'
)?
????????????
break
;
????????
else
????????????p
++
;
????}
????
return
?maxCount;
}
int
?main()?????
{??
????
char
?
*
p[
3
]
=
????
{
????????
"
String(1)8818882348888856788888888988888888880
"
,
????????
"
String(2)881888234888885678888888898888888880
"
,
????????
"
String(3)8818882348888856788888888988888888888888880
"
????}
;
????
int
?location?
=
?
0
;
????
char
?searchChar?
=
?
'
8
'
;
????
int
?tempCount?
=
?
0
;
????
int
?maxCount?
=
?
0
;
????
for
(
int
?i?
=
?
0
;i
<
sizeof
(p)
/
sizeof
(
*
p);i
++
)
????
{
????????tempCount?
=
?GetMaxCountofContinuousLetter(p[i],searchChar);
????????
if
(?tempCount?
>
?maxCount)
????????
{
????????????maxCount?
=
?tempCount;
????????????location?
=
?i;
????????}
????}
????cout
<<
"
the?string?of?maxCount?is?:?
"
<<
p[location]
<<
endl;
????cout
<<
"
the?count?of??the?continuous?letters?is?:?
"
<<
maxCount
<<
endl;
????cin.
get
();
}
/**/
/*
****************************************
*/
/**/
/*
?從str中刪除remove中含有的字符??????????
*/
/**/
/*
****************************************
*/
void
?RemoveChar(
char
?str[],?
char
?remove[])
{
????
int
?dst,src;
????
char
?removeArray[
256
];
????puts(str);
????puts(remove);
????
for
(src
=
0
;src
<
256
;src
++
)
????
{
???????removeArray[src]
=
0
;
????}
????src
=
0
;
????
while
(remove[src])
????
{
???????removeArray[remove[src
++
]]
=
1
;
????}
????src
=
0
;dst
=
0
;
????
while
(str[src])
????
{
????????
if
(
!
removeArray[str[src]])
????????
{
????????????str[dst
++
]
=
str[src];
????????}
????????src
++
;
????}
????str[dst]
=
0
;
????puts(str);
}
???? jeccy (燕飛月天) 于 2006年10月28日21:37:09 星期六 提到：
我的理解，給定2維數組
1 2 3三角形判斷
8 9 4
7 6 5
“螺旋式”順序輸出：1 2 3 4 5 6 7 8 9
【 在 bluexyz (數據結構是根本) 的大作中提到: 】
?
private
?
static
?
void
?PrintNumberasSpiralOrder()
????????
{
????????????
//
1??2??3???4
????????????
//
12?13?14??5
????????????
//
11?16?15??6
????????????
//
10?9???8??7
????????????
????????????
//
1,??2?,?3,??4,??5
????????????
//
16,?17,?18,?19,?6
????????????
//
15,?24,?25,?20,?7
????????????
//
14,?23,?22,?21,?8
????????????
//
13,?12,?11,?10,?9
????????????
int
[,]?a?
=
?
new
?
int
[,]
????????????
{?
{?
1
,?
2
,?
3
,?
4
,?
5
?}
,?
????????????
{?
16
,?
17
,?
18
,?
19
,?
6
?}
,?
????????????
{?
15
,?
24
,?
25
,?
20
,?
7
?}
,
????????????
{?
14
,?
23
,?
22
,?
21
,?
8
?}
,
????????????
{?
13
,?
12
,?
11
,?
10
,?
9
?}
?}
;
????????????
//
int[,]?a?=?new?int[,]?{?{?1,?2,?3,?4?},?{?12,?13,?14,?5?},?{?11,?16,?15,?6?},?{?10,?9,?8,?7?}?};
????????????
//
int[,]?a?=?new?int[,]?{?{?1,?2?},?{4,3}?};
????????????
const
?
int
?N?
=
?
5
;
????????????
//
?spiral
????????????
for
?(
int
?i?
=
?N?
-
?
1
,?j?
=
?
0
;?i?
>
?
0
;?i
--
,?j
++
)
????????????
{
????????????????
//
i?從最大維度數開始
????????????????
//
j?從最小維度數開始
????????????????
//
i--來縮小范圍
????????????????
//
j++來縮小范圍
????????????????
//
k?增長縮小因子
????????????????
for
?(
int
?k?
=
?j;?k?
<
?i;?k
++
)
????????????????????System.Console.Write(
"
{0}?
"
,?a[j,?k]);?Console.WriteLine(
""
);
????????????????
for
?(
int
?k?
=
?j;?k?
<
?i;?k
++
)
????????????????????System.Console.Write(
"
{0}?
"
,?a[k,?i]);?Console.WriteLine(
""
);
????????????????
for
?(
int
?k?
=
?i;?k?
>
?j;?k
--
)
????????????????????System.Console.Write(
"
{0}?
"
,?a[i,?k]);?Console.WriteLine(
""
);
????????????????
for
?(
int
?k?
=
?i;?k?
>
?j;?k
--
)
????????????????????System.Console.Write(
"
{0}?
"
,?a[k,?j]);?Console.WriteLine(
""
);
????????????}
????????????
//
?special?case?for?middle?element?if?N?is?odd
????????????
//
如果是奇數，則要條印中間的數。
????????????
if
?(N?
%
?
2
?
==
?
1
)?System.Console.Write(a[(N?
-
?
1
)?
/
?
2
,?(N?
-
?
1
)?
/
?
2
]);
????????}
Reference :
http://bigtiger2005.spaces.live.com/blog/cns!4e967c6fe38e7c14!141.entry?wa=wsignin1.0
摘要: 全排列的生成算法就是對于給定的字符集，用有效的方法將所有可能的全排列無重復無遺漏地枚舉出來。任何n個字符集的排列都可以與1～n的n個數字的排列一一對應，因此在此就以n個數字的排列為例說明排列的生成法。n個字符的全體排列之間存在一個確定的線性順序關系。所有的排列中除最后一個排列外，都有一個后繼；除第一個排列外，都有一個前驅。每個排列的后繼都可以從 它 的前驅經過最少的變化而得到，全排列的生成算法就是...
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?
void
?test10_2(
void
)
??
{
???
static
?
char
?
const
*
?szHexDigits?
=
?
"
01
"
;
???unsigned?
char
?b?
=
?
16
;
???cout?
<<
?szHexDigits[(b?
>>
?
7
)?
&
?
0x01
]?
<<
?szHexDigits[(b?
>>
?
6
)?
&
?
0x01
]?
<<
?
'
?
'
;
???cout?
<<
?szHexDigits[(b?
>>
?
5
)?
&
?
0x01
]?
<<
?szHexDigits[(b?
>>
?
4
)?
&
?
0x01
]?
<<
?
'
?
'
;
???cout?
<<
?szHexDigits[(b?
>>
?
3
)?
&
?
0x01
]?
<<
?szHexDigits[(b?
>>
?
2
)?
&
?
0x01
]?
<<
?
'
?
'
;
???cout?
<<
?szHexDigits[(b?
>>
?
1
)?
&
?
0x01
]?
<<
?szHexDigits[(b?
>>
?
0
)?
&
?
0x01
]?
<<
?
'
?
'
;
?}
?
void
?test10_16(
void
)
??
{
????
static
?
char
?
const
*
?szHexDigits?
=
?
"
0123456789abcdef
"
;???
???unsigned?
char
?b?
=
?
16
;
???cout?
<<
?szHexDigits[(b?
>>
?
4
)?
&
?
0x0f
]?
<<
?szHexDigits[b?
&
?
0x0f
]?
<<
?
'
?
'
;
?}